# Homework1 - Math408D Homework 1 Soultions Mike Harmon...

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Math408D Homework 1 Soultions Mike Harmon September 13, 2008 1 Section 7.8 18. lim x →∞ ln ( ln ( x )) x = 0 0 . So, use L’Hospitals rule. lim x →∞ ln ( ln ( x )) x = lim x →∞ d dx ( ln ( ln ( x ))) d dx ( x ) = lim x →∞ 1 ln ( x ) · 1 x 1 = 0 1 = 0 30. lim x 0 cos ( mx ) - cos ( nx ) x 2 = 0 0 L’Hospitals Rule lim x 0 cos ( mx ) - cos ( nx ) x 2 = lim x 0 d dx ( cos ( mx ) - sin ( nx )) d dx ( x 2 ) = lim x 0 - m · sin ( mx ) + n · sin ( nx ) 2 x = 0 0 Use L’Hospitals again! lim x →∞ cos ( mx ) - cos ( nx ) x 2 = lim x 0 d dx ( - m · sin ( mx ) + n · sin ( nx )) d dx (2 x ) = lim x 0 - m 2 · cos ( mx ) + n 2 · cos ( nx ) 2 = n 2 - m 2 2 1

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34. lim x x 2 + 2 2 x 2 + 1 Don’t use L’Hospitals Rule, it gets you no where with this problem. Instead divide both top and bottom by 1 x . lim x →∞ x 2 + 2 2 x 2 + 1 = lim x 1 x · x 2 + 2 1 x · 2 x 2 + 1 = lim x →∞ q 1 x 2 · ( x 2 + 2) q 1 x 2 · (2 x 2 + 1) = lim x →∞ q 1 + 2 x 2 q 2 + 1 x 2 = 1 2 38. lim x a + [ cos( x ) ln( x - a ) ln( e x - e a ) ] The trick is to remember that (lim x a f ( x ) g ( x )) = (lim x a f ( x )) · (lim x a g ( x )) So, with our problem this means that, lim x a + [ cos( x ) ln( x - a ) ln( e x - e a ) ] = lim x a + [cos( x )] · [ lim x a + ln( x - a ) ln(
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## This note was uploaded on 04/01/2009 for the course M 56435 taught by Professor Hamrick during the Spring '09 term at University of Texas at Austin.

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Homework1 - Math408D Homework 1 Soultions Mike Harmon...

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